 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  3 cents (Posted on 2014-08-29) Place two pennies on a table each touching a third, but not each other. The centers of the three form a certain angle.

For what angle will the area of the convex hull of this shape be maximized?

 No Solution Yet Submitted by Jer No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 3 of 3 | I found the same as Bractals, here is my method.
I called the radius of a penny 'r' and the convex angle θ.

The triangle of the centers of the pennies has sides:
2r, 2r, and 2 √(2) r (1-cos θ).

The area of the convex hull is the sum of the following areas:
* the central triangle formed by the centers of the pennies,
* 3 rectangles whose bases are each side of said triangle and with height r
* 3 arcs of a circle corresponding to the portions of the pennies not covered by either of the above (note that these 3 arcs add up to one full circle of radius r)

The area of the triangle is: 2 r� sin(θ)
The area of the 3 rectangles is:  4 � + 2 √(2) r� (1-cos θ)
The area of one circle:  πr�

Take the derivative wrt θ and set equal to zero:
2x� - 3x� + 1 = 0 (where x = cos(θ))
(x-1)� * (2x + 1) , so either cosθ = 1 or -(1/2)
So θ is either zero degrees (minimum area) or 120 degrees.
 Posted by Larry on 2014-08-31 11:08:23 Please log in:

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